A350987 Triangular numbers that are binary palindromes.

0, 1, 3, 15, 21, 45, 153, 231, 325, 561, 903, 2145, 4095, 8385, 14535, 33153, 58311, 63903, 101475, 131841, 240471, 399171, 525825, 932295, 976503, 1044735, 1308153, 1395285, 1532125, 2100225, 3727815, 8394753, 14680071, 17913105, 33566721, 54054003, 59650503

Offset: 1

Author

Amiram Eldar, Jan 28 2022

Keywords

• nonn
• base

Comments

This sequence is infinite since (2^k+1)*(2^k+2)/2 = A000217(A028401(k)) is a term for all k>1 (Trigg, 1974).

Examples

			3 is a term since 3 = A000217(2) = 2*(2+1)/2 is a triangular number, and 3 = 11_2 is also a binary palindromic number.
15 is a term since 15 = A000217(5) = 5*(5+1)/2 is a triangular number, and 15 = 1111_2 is also a binary palindromic number.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..128
• Charles W. Trigg, Infinite sequences of palindromic triangular numbers, The Fibonacci Quarterly, Vol. 12, No. 2 (1974), pp. 209-212.
• Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT], 2008.

Crossrefs

Intersection of A000217 and A006995.

The binary version of A003098.

A028401 \ {6} is a subsequence.

Cf. A350988, A350990 (base 3), A350991, A350992, A350993.

Programs

Mathematica

t[n_] := n*(n + 1)/2; Select[t /@ Range[0, 2*10^4], PalindromeQ[IntegerDigits[#, 2]] &]

Python

from itertools import count, islice
def agen():
    for i in count(0):
        t = i*(i+1)//2
        b = bin(t)[2:]
        if b == b[::-1]:
            yield t
print(list(islice(agen(), 37))) # Michael S. Branicky, Jan 28 2022

Formula

a(n) = A000217(A350988(n)).